Question

Xenon crystallizes in fcc lattice and the edge length of unit cell is 620 pm. What is the radius of Xe atom?

• A. 219.2 pm
• B. 438.5 pm
• C. 265.5 pm
• D. 536.9 pm

Hint:

For fcc lattices, remember the relation \(a = 2\sqrt{2}r\) to calculate the atomic radius from the edge length.

Answer 1

The Correct Option is A

Step 1: Understanding fcc lattice.
In a face-centered cubic (fcc) lattice, the relationship between the edge length \(a\) and the atomic radius \(r\) is given by: \[ a = 2\sqrt{2} \, r \] Where \(a = 620 \, \text{pm}\) is the edge length of the unit cell. We can rearrange the equation to solve for \(r\): \[ r = \frac{a}{2\sqrt{2}} = \frac{620}{2\sqrt{2}} = 219.2 \, \text{pm} \]
Step 2: Conclusion.
The radius of the xenon atom is 219.2 pm.